Grade 2 Math 3-digit Addition With Regrouping
3-digit Addition With Regrouping
Here we shall learn how to
add 3-digit numbers with regrouping or carry forward in detail. 3-digit Addition With Regrouping is solved in
three different ways-
1. Expanded Method in words
a. Add together 237, 378 and 333 using expanded method in words.
Solution:
By expressing the numbers in expanded form in words, we get
237 = 2 hundreds + 3 tens + 7
ones
378 = 3 hundreds + 7 tens + 8 ones
= 8 hundreds + 13 tens + (1 ten + 8 ones)
= 8 hundreds + 14 tens + 8 ones
= 8 hundreds + (1 hundred + 4 tens) + 8 ones
= 9 hundreds + 4 tens + 8 ones
Explanation:
Adding one’s place digits = 7 + 8 + 3 = 18 ones = 1 tens + 8 ones … (i)
Adding ten’s place digits = 3 + 7 + 3 = 13 tens = 1 hundred + 3 tens … (ii)
Adding hundred’s place digits = 2 + 3 + 3 = 8 hundreds … (iii)
As per the carry over process we shall add the results of (i), (ii) and (iii). Here, we shall add same placed digits only.
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1 tens
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8 ones
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1 hundred
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3 tens
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8 hundreds
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-------------------------------------------
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9 hundreds
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4 tens
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8 ones
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= 948 (Ans)
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2. Expanded Method in Numbers
a. Add together 286, 278 and 364 using expanded method in figures.
Solution:
By expressing the numbers in expanded form in numbers, we get
286 = 200 + 80 + 6
278 = 200 + 70 + 8
= 700 + (200 + 10) + (10 +
8)
= 700 + 200 + 20 + 8
= 900 + 20 + 8
= 928
Explanation:
Adding one’s place digits = 6 + 8 + 4 = 18 ones = 10 + 8 … (i)
Adding ten’s place digits = 8 + 7 + 6 = 21 tens = 210 = 200 + 10 … (ii)
Adding hundred’s place digits = 2 + 2 + 3 = 7 hundreds = 700 … (iii)
As per the carry over process we shall add the results of (i), (ii) and
(iii). Here, we shall add same placed digits only.
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10
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8
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200
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10
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700
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-------------------------------
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900
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20
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8
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= 928 (Ans)
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3. Short Method
a. Add together 426, 543 and 298 in short method.
Solution:
By Short Method:
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4
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2
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6
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5
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4
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3
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2
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9
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8
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------------------------
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Sum
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11
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15
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17
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=
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11
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16
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7
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=
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12
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6
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7
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Explanation:
Adding one’s place digits = 6 + 3 + 8 = 17 ones = 1 ten + 7 ones … (i)
Adding ten’s place digits = 2 + 4 + 9 = 15 tens = 1 hundred + 5 tens … (ii)
Adding hundred’s place digits = 4 + 5 + 2 = 11 hundreds … (iii)
As per the carry over process we shall add the results of (i), (ii) and (iii). Here, we shall add same placed digits only.
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1 ten
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7 ones
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1 hundred
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5 tens
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11 hundreds
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------------------------------------------
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12 hundreds
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6 tens
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7 ones
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= 1267 (Ans)
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Read More:
2-digit Addition With Regrouping
Conclusion:
All method's results are the same. So learn all methods, and practice yourself more and more. Share this post and go ahead.